LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (openai/gpt-5.4)verified 2026-07-26 (claude-opus-5)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Everything in $M$ is comparable to an extremal element

Statement

Let $(P, \leq)$ be a chain-complete poset, $f : P \to P$ progressive, $M$ the smallest admissible set, and $x \in M$ extremal (Extremal element and its cut (Bourbaki–Witt)). Then $M_{x} = M$ ; that is, for every $y \in M$ , either $y \leq x$ or $f (x) \leq y$ .

Facts & Assumptions

Given: A chain-complete poset $(P, \leq)$ , a progressive $f : P \to P$ , the smallest admissible set $M$ , and an extremal $x \in M$ .

[A1]

$M_{x} = {z \in M : z \leq x or f (x) \leq z}$ , so $M_{x} \subseteq M$ by construction (Extremal element and its cut (Bourbaki–Witt)).

[L1]

$M_{x}$ is closed under $f$ (The cut at an extremal element is closed under $f$ ).

[L2]

$M_{x}$ is closed under suprema of its chains (The cut at an extremal element is closed under chain suprema).

[L3]

$M$ is contained in every admissible subset of $P$ (A smallest admissible set exists).

[L4]

A subset is admissible when it is closed under $f$ and under suprema of its chains (Admissible subset (Bourbaki–Witt)).

Proof

technique · direct

1.1L1

$M_{x}$ is closed under $f$ .

1.2L2

$M_{x}$ is closed under suprema of its chains.

2.1step 1.1step 1.2L4

So $M_{x}$ is an admissible subset of $P$ .

3.1step 2.1L3

By minimality of $M$ , every admissible subset contains $M$ , so $M \subseteq M_{x}$ .

4.1step 3.1A1

Together with $M_{x} \subseteq M$ this gives $M_{x} = M$ .

5.1step 4.1A1∎

Unfolding the definition of $M_{x}$ , every $y \in M$ satisfies $y \leq x$ or $f (x) \leq y$ .

Remarks

This is the first payoff of minimality, and the pattern is worth naming: to prove that everything in $M$ has a property, collect the elements that have it, show the collection is admissible, and let minimality do the rest. The same move proves Every element of $M$ is extremal.
The conclusion is a comparability statement with a gap. It says nothing about elements strictly between $x$ and $f (x)$ , and indeed the content of the lemma is that $M$ has none: an element of $M$ lies at or below $x$ , or at or above $f (x)$ , never inside.
The hypothesis that $x$ is extremal is doing real work and cannot be dropped. It is what Every element of $M$ is extremal later supplies for every element, which is what turns this one-sided statement into total comparability (The smallest admissible set is a chain).

Everything in $M$ is comparable to an extremal element

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Everything in MM is comparable to an extremal element

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Everything in $M$ is comparable to an extremal element